Question: $h(t) = (4t^2, -5\cos(t^2))$ What is the velocity of $h(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{64 + 400t^4\cos^2(t^2)}$ (Choice B) B $(8, 20t^2 \cos(t^2))$ (Choice C) C $\sqrt{64t^2 + 100t^2\sin^2(t^2)}$ (Choice D) D $(8t, 10t \sin(t^2))$
Explanation: The velocity of a parametric curve is the derivative of its position. If $f(t) = (a(t), b(t))$, then the velocity is: $f'(t) = (a'(t), b'(t))$ Our position function here is $h(t)$. $h'(t) = (8t, 10t \sin(t^2))$ Therefore, the velocity of $h(t)$ is $(8t, 10t \sin(t^2))$.